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Let $E\subset \mathbb{R}$ be a Lebesgue measurable subset of reals such that $\mu(E)>0.$ Consider the set $E+E=\{x+y: x,y\in E\},$ prove that $E+E$ contains an interval of length greater than $0$.

Is this problem even true? Let's say the set $E$ is our set of irrational numbers, then $E+E$ doesn't contain positive length interval. If this is indeed true, how can we go about proving this statement?

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    $\begingroup$ This is an exiting statement and i don't know if it is true. The irrationals aren't a counterexample however. If $x\in E$, then $x = \frac{x}{2}+\frac{x}{2} \in E+E$. If $x\in\mathbb Q$, then $(x-\pi) + \pi \in E+E$. Hence $E+E = \mathbb R$ in this case. $\endgroup$
    – Lukas Betz
    May 11, 2015 at 1:00
  • $\begingroup$ Hint: The function $y\to \int_E\chi_E (y-x)\,dx$ is continuous. $\endgroup$
    – zhw.
    May 11, 2015 at 1:24
  • $\begingroup$ In the first paragraph, do you mean contains instead of consists? $\endgroup$
    – Michael Burr
    May 11, 2015 at 1:25
  • $\begingroup$ @MichaelBurr: Yes. sorry for typo. $\endgroup$
    – James Bond
    May 11, 2015 at 1:31

2 Answers 2

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This is true in general, for sets $A,B\subseteq\mathbb R$ with $\mu(A),\mu(B)>0$.

A way to do this is to consider the characteristic functions $\chi_A,\chi_B$, and set $f=\chi_A*\chi_B$ to be their convolution. Then $f$ is nonnegative continuous, and we can compute that $\int_{\mathbb R}f=\mu(A)\mu(B)>0$, so, from continuity, $$\int_{\mathbb R}\chi_A(y)\chi_B(x-y)\,dy\geq\varepsilon$$ for all $x$ in an interval $(a,b)$. This shows that there exists $y\in\mathbb R$ such that $\chi_A(y)\chi_B(x-y)>0$, therefore $y\in A$ and $x-y\in B$. This will show that $x\in A+B$, therefore $(a,b)\subseteq A+B$.

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  • $\begingroup$ Do you mean $f= \chi_A * \chi_B= u(A)u(B)$ or by $\int_R f$ do you mean $\int \int \chi_A(x-y) \chi_B(y) dy dx = u(A) u(B)$? I can show by translation invariance that the double integral does indeed equal u(A) u(B) but not the first integral.. If you do mean the second integral, then how do you conclude that $ \chi_A * \chi_B \geq \epsilon$ since we showed the integral of f is strictly positive (ie the double integral) and not f itself(the single integral) ? Please clarify.. $\endgroup$
    – user172377
    Oct 10, 2016 at 7:56
  • $\begingroup$ $f$ is the convolution of $\chi_A$ and $\chi_B$; indeed, the double integral is equal to $\mu(A)\mu(B)$. Since now $f$ is continuous and its integral is positive, it is positive at some point $x_0$. Then, for $x$ sufficiently close to $x_0$, $f(x)\geq\varepsilon$, for some positive $\varepsilon$. $\endgroup$
    – detnvvp
    Oct 13, 2016 at 6:59
  • $\begingroup$ thanks! basically since f is non negative and its integral is positive, it is positive almost everywhere, so then we're choosing one of those $x_0$ where it is positive and deducing from the continuity that for x sufficiently close, $f(x)\geq \epsilon$ ? $\endgroup$
    – user172377
    Oct 13, 2016 at 19:22
  • $\begingroup$ Yes, this is correct. $\endgroup$
    – detnvvp
    Oct 15, 2016 at 4:13
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First, let me give some intuition on the idea to be used: If the set $E$ were finite, an idea to "detect" whether a certain point $x$ belongs to $E+E$ would be to look at $\chi_{E}(x-y)$ for every $y \in E;$ that is, to study

$\sum_{y \in E}\chi_{E}(y)\chi_{E}(x-y)$

In the continuous case, we could try something similar by studying the convolution. Consider thus the function:

$\phi(x)=\chi_{E} \ast \chi_{E}(x).$

Notice that it is continuous, as you can easily check by using the dominated convergence theorem. Then the set $A=\{x \in \mathbb{R} \ : \ \phi(x)>0 \}$ is open, which means that it must contain an open interval.

It would be enough then to check that $A \subset E+E.$ But this actually obvious, so we are already done.

EDIT. Sorry, I didn't see that another answer was posted... it happened while I was writing. Let me know if I should delete mine.

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  • $\begingroup$ Sorry, @Qwertuy, but what do you mean that you can use the DCT to prove continuity of $\phi$? $\endgroup$
    – rem
    Mar 6, 2019 at 1:11
  • $\begingroup$ it is a pretty trivial statement. I had in mind approximating the $\chi_E$ by smooth functions and bounding the convolution by convolutions of these smooth functions, but of course this is all superflous, since the continuity of the convolution was already evident. $\endgroup$
    – Qwertuy
    May 5 at 14:06

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